Worked solution · Exam P

SOA Exam P Sample Question #319 Solution

This Exam P sample reference tests Independence. Complement the requested event by calculating exactly three purchases and exactly four purchases. Those probabilities are 0.3626 and 0.0686, so the complement is 0.5688 and choice C.

Independent analysis by ActuaryProof · reviewed July 31, 2026

Official wording withheld. Read the problem in the official SOA sample-question PDF. This page provides only our independently written analysis.

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Independent solution

How to solve this Independence question

Setup

Let K be the total number of purchases. It is shorter to complement the two counts above two.

\Pr(K\le2)=1-\Pr(K=3)-\Pr(K=4)

Model

Exactly three purchases can occur with the fourth customer purchasing and exactly two of the first three purchasing, or with all first three purchasing and the fourth not purchasing.

\Pr(K=3)=\binom32(0.7)^2(0.3)(0.2)+(0.7)^3(0.8)

=0.0882+0.2744=0.3626

Compute

Calculate all four purchases and subtract both upper-count probabilities.

\Pr(K=4)=(0.7)^3(0.2)=0.0686

\Pr(K\le2)=1-0.3626-0.0686=0.5688

Answer

The probability rounds to 0.57.

\boxed{0.57\quad\text{(C)}}

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Error diagnosis

Why the other letters tempt

These notes identify the calculation error associated with each wrong letter when that error is reproducible.

AThis is P(K≤1)=0.0216+0.1566=0.1782, which rounds to 0.18. It excludes the allowed two-purchase outcomes.
BThis is P(K=2)=0.3906, only one of the three count outcomes included in the phrase at most two.
DThis cannot be a correct cumulative count probability: even if the fourth purchase probability were zero, P(K≤2) for the first three would be only 1-0.7³=0.657. It necessarily includes part of a three-purchase outcome.
EThis is P(K≥2)=1-P(K≤1)=1-0.1782=0.8218. It reverses the requested lower-count event.

Original practice · fully worked

Original variant: automatic control with manual fallback

A control unit has independent components. Its controller works with probability 0.68, and two sensors work with probabilities 0.73 and 0.57. When the controller works, the unit operates if at least one sensor works. When the controller fails, the unit instead operates only if an independent manual switch works; that switch has probability 0.46 of working. Calculate the probability that the unit operates.

A
0.147
B
0.601
C
0.748
D
0.785
E
0.884

Answer and calculation

The automatic and manual branches are disjoint because they require opposite controller states. Their probabilities are 0.601052 and 0.1472, which sum to 0.748252 and choice C.

Setup

Begin with the complementary chance that neither independent sensor is available.

p_S=1-(1-0.73)(1-0.57)=1-(0.27)(0.43)=0.8839

Model

Partition operation into disjoint automatic and manual branches according to the controller state.

p_A=0.68(0.8839)=0.601052

p_M=(1-0.68)(0.46)=0.1472

Compute

Add the two disjoint branch probabilities.

\Pr(\text{operates})=p_A+p_M

=0.601052+0.1472=0.748252

Answer

The operating probability rounds to 0.748.

\boxed{0.748\quad\text{(C)}}

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How to solve this Independence question

Setup

Model

Compute

Answer

General Probability Practice Workbook

Why the other letters tempt

Original variant: automatic control with manual fallback

Answer and calculation

Setup

Model

Compute

Answer

Related Exam P solutions

Next steps for Exam P General Probability